# Fibration on the category of Lie pseudoalgebras implementing comorphisms

I am trying to understand comorphisms of Lie pseudoalgebras from the point of view of fibred categories, but failing miserably so far. My question would be:

Is there a (op)fibration $$\mathrm{LiePs} \to \mathrm{Alg}$$ of the category of Lie pseudoalgebras in the sense of (op)fibred categories such that comorphisms of Lie pseudoalgebras are morphisms in the dual fibration $$\mathrm{LiePs}^*$$?

Let me be a bit more precise: Lie pseudoalgebras (aka Lie-Rinehart algebras or others) can be seen as the algebraic counterparts of Lie algebroids.

Definition: (Lie Pseudoalgebra)
A Lie pseudoalgebra consists of a commutative algebra $$\mathcal{A}$$ and a Lie algebra $$\mathfrak{g}$$, such that $$\mathfrak{g}$$ is an $$\mathcal{A}$$-module and $$\mathcal{A}$$ acts on $$\mathfrak{g}$$ by derivations, i.e. we have a Lie algebra morphism $$\rho \colon \mathfrak{g} \to \mathrm{Der}(\mathcal{A})$$ which is also an $$\mathcal{A}$$-module morphism and we have $$[\xi_1, a \xi_2] = a[\xi_1,\xi_2] + \rho(\xi_1)(a)\xi_2$$ for $$\xi_1, \xi_2 \in \mathfrak{g}$$, $$a \in \mathcal{A}$$.

For Lie pseudoalgebras there is a quite obvious notion of morphism.

Definition: (Morphism of Lie pseudoalgebras)
A morphism of Lie pseudoalgebras $$(\mathcal{A},\mathfrak{g})$$ and $$(\mathcal{B},\mathfrak{h})$$ consists of an algebra morphism $$\phi \colon \mathcal{A} \to \mathcal{B}$$ and a module morphism $$\Phi \colon \mathfrak{g} \to \mathfrak{h}$$ along $$\phi$$ which is also a Lie algebra morphism and $$\phi(\rho_{\mathcal{A}}(\xi)a) = \rho_{\mathcal{B}}(\Phi(\xi))\phi(a)$$ holds for $$\xi \in \mathfrak{g}$$, $$a \in \mathcal{A}$$.

Let us denote the category of Lie pseudoalgebras by $$\mathrm{LiePs}$$. Then there is an obvious functor $$\mathrm{LiePs} \to \mathrm{Alg}$$ by mapping a Lie pseudoalgebra $$(\mathcal{A},\mathfrak{g})$$ to the algebra $$\mathcal{A}$$ and a morphism $$(\Phi,\phi)$$ to the corresponding algebra morphism $$\phi$$.

Question 1: Is $$\mathrm{LiePs} \to \mathrm{Alg}$$ a (op)fibration of categories?

If so: for (op)fibred categories there is a notion of a dual (op)fibration, see e.g. A.Kock; The dual fibration in elementary terms, whose morphisms can be understood as comorphisms of the original objects. And there is a notion of comorphism of Lie pseudoalgebras, see e.g. Z. Chen,Z.-J. Liu; On (co-)morphisms of Lie pseudoalgebras and groupoids.

Question 2: Do comorphisms of Lie pseudoalgebras agree with morphisms in the dual fibration $$\mathrm{LiePs}^* \to \mathrm{Alg}$$?

## 1 Answer

Not a direct answer but this is how I would tackle this:

Background knowledge: Given a category $$C$$ and a (weak-) functor $$F:C^{op}\to \mathrm{Cat}$$ one can construct a category $$\int F$$ (also denoted $$\int_C F$$) called the Grothendieck Construction that comes with a canonical arrow $$\pi_f:\int F\to C$$ that happens to be a fibration. Inversely: Given a fibration $$\pi: E\to C$$ we can construct a (weak-) functor $$F_\pi:C^{op}\to \mathrm{Cat}$$. These two constructions are inverse in a suitable way. Similarly for covariant functors $$F:C\to\mathrm{Cat}$$.

Now: To check if your functor $$\mathrm{LiePs}\to\mathrm{Alg}$$ is a fibration I would try to construct $$\mathrm{LiePs}$$ as the Grothendieck Construction of a functor $$Ps:\mathrm{Alg}^{op}\to\mathrm{Cat}$$ (or maybe $$Ps:\mathrm{Alg}\to\mathrm{Cat}$$).

The object part of this functor should be

$$Ps: A \mapsto \mathrm{Lie}^*_{/\mathrm{Der}(A)}$$ where $$\mathrm{Lie}^*_{/\mathrm{Der}(A)}$$ is a suitable subcategory of the comma category $$\mathrm{Lie}_{/\mathrm{Der}(A)}$$. A nice feature of the Grothendieck Construction is the following: If your functor actually is a fibration, you should then be able to extend this assignment to a weak functor in a more or less obvious way: For every morphism $$f:A\to B$$ in $$\mathrm{Alg}$$ you would search for a functor $$f^*:\mathrm{Ps}(B)\to \mathrm{Ps}(B)$$ (Or maybe covariantly $$f_*:\mathrm{Ps}(A)\to \mathrm{Ps}(B)$$?).

The next step is to check if the resulting category $$\int Ps$$ is the "same" as $$\mathrm{LiePs}$$. From the construction of $$\int Ps$$ there should already by suitable candidates for the equivalence functors.